Let A be a n x n matrix such that A^k = 0 (k some natural number).
Prove that A^n = 0.
Any ideas?
Thanks.
Let v be an eigenvector. Then Av=cv, where c is the corresponding eigenvalue. A^k v=c^k v=0. Then c must be 0. Therefore, there is only one eigenvalue. Now J=P^-1 A P, where J is the Jordan normal form of the matrix and P is a non singular matrix. You can easily check that J^n=P^-1 A^n P. Now, the Jordan normal form is either 0, in which case, Ju=P^-1 A P u=0, for any vector u, which implies that A is 0; or the Jordan normal form is a matrix with all entries 0, except above the diagonal where they are 1's (some authors prefer to put the 1's below the diagonal). Now you can also check that after n multiplications of any vector u with J^n, it will be 0, which implies that A^n is 0.
the fact that A^{k} = 0, means that the minimal polynomial for A, m(x), divides x^{k} (*), so is of the form x^{t}, for some t.
the fact that every root of the characteristic polynomial (eigenvalue) is also a root of the minimal polynomial (**), means that the characteristic polynomial of A is:
(x - 0)^{n} = x^{n}.
since every matrix A satisfies its characteristic polynomial (Cayley-Hamilton), A^{n} = 0.
a proof of (*):
suppose that p(A) = 0, for some polynomial p(x), and let m(x) be the minimal polynomial for A.
writing p(x) = q(x)m(x) + r(x), where either r(x) = 0, or deg(r) < deg(m), we have:
0 = p(A) = q(A)m(A) + r(A) = q(A)(0) + r(A) = r(A).
if deg(r) < deg(m), this contradicts the minimality of m, so we must have r(x) = 0 (the 0-polynomial), so that p(x) = q(x)m(x), so m(x) divides p(x).
a proof of (**):
suppose λ is an eigenvalue of A (and thus a root of det(xI - A)), so that for some eigenvector v: A(v) = λv, and let the minimal polynomial of A be m(x), so that m(A) = 0 (the 0-matrix).
then (m(A))(v) = (m(λ))v. but (m(A))v = 0v = 0, so that m(λ) must be 0, that is, λ is a root of m(x).
yes! the JNF of such a matrix must consist of 0's on the diagonal, and some number of 1's on the superdiagonal. so:
A = PNP^{-1}, where N is of nilpotency ≤ n. whence:
A^{n} = PN^{n}P^{-1} = P(0)P^{-1} = 0.
(here, my "N" is your "J"...i typically write J = D+N, where D is diagonal and N is nilpotent. the first part of your post shows that D = 0).
postscript: my answer was in no way intended to invalidate ModusPonens' excellent reply. i wanted to illustrate the richness of linear algebra: often, there are several ways to look at (and solve) a given problem.